POJ 2393 Yogurt factory

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Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13459		Accepted: 6753

Description

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week. 

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt. 

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S. 

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

Hint

OUTPUT DETAILS: 
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units. 

Source

USACO 2005 March Gold

#include 
#include 
#include 
#include 
#include 

using namespace std;
long long i,j,n,s,k,ans,c[20000],y[20000],a[20000],sum[20000],min1;
int main()
{
    scanf("%lld%lld",&n,&s);
    for (i=1;i<=n;i++)
    scanf("%lld%lld",&c[i],&y[i]);
    for (i=1;i<=n;i++)
        sum[i]=c[i]-s*i;
    min1=0x3f3f3f3f;
    k=0;
    for (i=1;i<=n;i++)
    {
        if (sum[i]>min1) a[i]=k; else a[i]=i;
        if (sum[i]

PS：当前week i的果汁总共有两种操作，一个是这周做好，一个是前面预先做好，到底选择哪个方案呢？

1.如果本周做好，花费为c[i]*y[i]；

2.如果在前面week j做好，肯定是在前面week j先做好比在本周做更优的情况下，也就是满足c[j]*y[i]+(j-i)*s*y[i]，我们选择j（j<=i）中使得c[j]*y[i]+(j-i)*s*y[i]最小的一个。

如果采用暴力手段，会超时。

这时我们注意到式1两边可以同时约去y[i]，式1就变成了c[j]+(j-i)*s

做题不要慌，慢慢来，祝大家刷题顺利！加油！